1. ## Chain rule question...

Given the function:
$\displaystyle s(t)=\sin2\pi*t$

My book says: $\displaystyle s'(t)=2\pi*\cos2\pi*t$

I'm pretty sure this is a utilization of the chain rule, but I don't really understand it, can someone clear it up for me?

(ps, sorry I could not get it to show up any clearer...)

2. chain rule is g'(h(x)) h'(x) . which means..

g(x)=sin(x) [x being sin(2Pi)]
g'(x)=cos(x)
h(x)=sin(2Pi) and
h'(x)=0 .

put it all into the function and you get..

cos(sin(2Pi)) (0)

which ends up giving you a 0 anyways. are you sure thats what the book is saying? lol i mean it sounds dumb to ask but this is the way i did it.. idk and i tried many times on the calculator and i keep getting zero because the derivative of t is 1.. and anything that you multiply by zero stays zero... you know??? any more info you have?

3. Originally Posted by rust1477
Given the function:
$\displaystyle s(t)=\sin2\pi*t$

My book says: $\displaystyle s'(t)=2\pi*\cos2\pi*t$

I'm pretty sure this is a utilization of the chain rule, but I don't really understand it, can someone clear it up for me?

(ps, sorry I could not get it to show up any clearer...)
Is s(t) = sin(2pi times t), or

sin(2pi) times t?

4. yea! lol.. im pretty sure thats it.. that makes more sense. hehe how come i didnt see that?? good job.

5. Originally Posted by rust1477
Given the function:
$\displaystyle s(t)=\sin2\pi*t$

My book says: $\displaystyle s'(t)=2\pi*\cos2\pi*t$

I'm pretty sure this is a utilization of the chain rule, but I don't really understand it, can someone clear it up for me?

(ps, sorry I could not get it to show up any clearer...)
Clearly the function is $\displaystyle s(t) = \sin (2 \pi t)$. Please post your equations so that they are less ambiguous.

You're expected to know the standard form: If $\displaystyle y = \sin (kt)$ then $\displaystyle \frac{dy}{dt} = k \cos (kt)$.